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Thermodynamics I Gibbs equation, Lecture notes of Thermodynamics

using gibbs equation to solve thermodynamic problems

Typology: Lecture notes

2020/2021

Uploaded on 02/01/2021

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HW9 worksheet due today
Handout HW10
First task will be to gain a conceptual understanding of
the 2nd law
Second task will be to develop a physical interpretation
of entropy
using the 2nd law
Goal with 2nd Law Lectures
Develop Gibbs (i.e. TdS) Equations… PPT
Begin 2nd law for a closed system
Today:
ME 291 L21 11
-
16
-
16
Wednesday, November 16, 2016 12:40 PM
ME 291
-
F16 Page 1
pf3
pf4
pf5
pf8
pf9

Partial preview of the text

Download Thermodynamics I Gibbs equation and more Lecture notes Thermodynamics in PDF only on Docsity!

• HW9 worksheet due today

• Handout HW

First task will be to gain a conceptual understanding of

the 2nd law

Second task will be to develop a physical interpretation

of entropy

Third task will be to do more problems, but this time

using the 2nd law

• Goal with 2nd Law Lectures

• Develop Gibbs (i.e. TdS) Equations… PPT

• Begin 2nd law for a closed system

• Today:

ME 291 L21 11- 16 - 16

Wednesday, November 16, 2016 12:40 PM

ME 291 - F16 Page 1

ME 291 – Thermodynamics I

The “Gibbs Equations” a.k.a. the “Tds Equations”

Tom Acker, PhD

Professor of Mechanical Engineering College of Engineering, Forestry and Natural Sciences

1

MechanicalEngineering

2

MechanicalEngineering

^

Thermal equilibrium

  • When two objects in thermal

contact with one another reach the sametemperature and heat transfer no longer occursbetween the objects.

^

Temperature is our measure of the potential for heattransfer between two objects

. When two objects in

thermal contact no longer exchange energy as heat,we say the objects are at the same temperature

^

When heat transfer no longer occurs between twoobjects in thermal contact and the

system is in

equilibrium, then the entropy is maximized for theprocess

(via the second law of thermodynamics)

Thermal Equilibrium

WarmObject

CoolObject

ThermalEquilibrium

Time

Both objects atsame temperature

3

MechanicalEngineering

^

In this description of thermal equilibrium:^ ^

The

temperature difference drives the energy transfer as heat

between the objects ^ The

temperature is an indication of the internal energy

of

the objects  The

temperature is related to the entropy

. The more

internal energy, the more ways to store the energy (morepossible macrostates and microstates,

w ), the more

entropy (

S = k ln w

^ S is maximized at thermal equilibrium

via the 2

nd^ Law for

an isolated system (microscopic: most likely macrostate) ^

Let’s relate T to S and U (Note, T is directlymeasurable)

WarmObject

CoolObject

ThermalEquilibrium

Time

T, S, & Thermal Equilibrium

4

MechanicalEngineering

constant

2 1

  

^

B A^

U U U

U U

(^

2 2

1 1

B A

B A^

U

U

U

U^

^

B B A A^

U

U

U

U

here

2 1 2 1

w^

0   E

^

st 1

Law, isolated system:

) ,

neglect( 0

0

PE KE

U

E^

      

^ Note, the amount of energy on each side changes as heat istransferred between A & B. On a differential basis: Simplify with 1

st^

Law:

^

B A^

dU

dU

dU

B A^

dU dU

^

The change of energy in A andB are equal but opposite

9

)

( )

( as

Rewrite

1 2

1 2

B B

A A^

U U

U U^

   

MechanicalEngineering

B

A^

B B

A A

S U

S U

    0

B

A^

B B

A A

S U

S U

This holds atthermal equilibrium

Plug (2) into (1):

^

A

B B A

A^ A

dU

S U

dU S U dS

B

A

  

 

^

At thermal equilibrium, the entropy is maximized andno longer changing. Therefore

dS = 0

 



     

^

B

A^

B B

A A A^

S U

S U

dU dS^

0

10

MechanicalEngineering

v

v^

u T c^

c^ v^

is a property of a substance and is defined in terms of other properties (

u ,^ T ,and

v )

^  

S U

^ 

S U

Compare to Specific Heat ^ Furthermore,

^ is a property that has the same value for twosubstances in thermal equilibrium; ^ since temperature is the same for two substances inthermal equilibrium, this new property must be related tothe temperature ^ Recall the definition of

c^ v

:

^ Similarly,

is a property of a substance

11

MechanicalEngineering B

B B A

A^ A

dU S U

dU S U dS

B

A^

B

A^

dU

dU

Recall Eq. (1):For any real process,

dS

0 (irreversible process),

And Eq. (2):Plug (2) into (1):

(^0)    

  

   

^

B

A^

B B

A A A^

S U

S U

dU dS

Consider Any Process:

Let’s relate

dU

to the heat transfer between A & B A^

12

MechanicalEngineering

A

A

A^

W

Q

dU

 ^

Now write the first law for side A:

0 Heat transfer into A from B

Plug into (4):

^

B

A^

B B

A A A^

S U

S U

Q

dS

^

SideA (5)

Define Side A as the System

Q ^ A W ^ A

dS

is always positive If^ δ

QA

0, then the term in brackets must be > This occurs only when

TA

<^ TB

and

B

A^

B B

A A

S U

S U

MechanicalEngineering

^ Since

when

;^ inverselyrelated

An appropriate relationship between

T ,^ S

, and

U^

is: Plug (6) into (4), and check sign on

dS

:

B

A^

B B

A A

S U

S U

B A^

T T^

  S ^ UT^

1

T S^ U

1

or^

  

Thermodynamic Definition ofTemperature

^ 

B A A^

T

T

Q

dS

^

If^ T

<TA

, then B

δQ

0, and A^

dS

is +

If^ T

>TA

, then B

δQ

< 0, and A^

dS

is +

MechanicalEngineering

^ For the units of Eq. (6) to be consistent,entropy must have units of energy pertemperature: SI^

R

Btulbm s BtuR S^

^

K

kJkg s kJK S^

Check Units on S

T

S^ U

e temperatur

energy

Sof units

1 ^ 



e energytemperatur Sof units

English

So, our definition

makes sense

, is

dimensionally consistent

and satisfies the requirement that

S is maximized

at thermal

equilibrium

15

MechanicalEngineering

16

MechanicalEngineering

^

To find max, take derivative and set equal to 0,

dS=

Recall:

Maximize Entropy

(^0)        

  

 

B B BU

A AU A

B B B

A

A^ A

d S

d S

dU S U

dU S U dS

B

A

B

A

B A^

dS

dS

dS

& ) , (^

^

US S From geometry:From 1

st^ Law:

A

B^

dU

dU

A

B^

d

d^

(^0)  



       



     

^

B

A

B

A^

BU B

AU A A

B B

A A A

S

S

d

S U

S U

dU dS

21

MechanicalEngineering

Recall Eq (6):

So this simplifies to

T S^ U

1   

^ 

B

A^

U B B

U A A A

B A A

S

S

d

T

T

dU

dS

B A^

T T^ 

^

sincem

equilibriuat 0

B

A^

U B B

U A A A

S

S

d

dS^

Simplify

22

MechanicalEngineering

^

Thus, in this example we have

another property

that we must be equivalent for A and B when thesystems is in equilibrium:^ ^

Thermal equilibrium (

TA^

= T

) B

^ Mechanical equilibrium (

p^ A^

=^ p

)B

B

A^

U B B

U A^ A

S

S

U S^  

(^0)  

B

A^

U B B

U A A A

S

S

d dS^ ^

Equals zero at equilibrium if:

^

This new property is

intensive

, just as

T^

and

p^

are.

^

Let’s relate this new property to

T^

and

p.

23

MechanicalEngineering

S 

volume

e

temperatur energy

3 lengthe

temperatur

length force

^

e

temperatur

length force

2

^

e

pressuretemperatur

U S^  

New Property:^ ^

Check units on our new property:

P T

S

U   

^

Since the units of this property are pressure dividedby temperature, let’s define it as follows and see if itmakes sense:

24

MechanicalEngineering (^0)  

B

A^

BU B

AU A A

S

S

d dS

^ 

B B A A A^

P T

P T

d dS

^

^

^

B A A^

P

P

d T dS

Check if it Makes Sense^ ^

Plug into Eq. (7):

^

Assume for the moment that

TA

=^

TB

, then

^ If p

< A^

p^ B

, then

( p

– A^

p^ B

) < 0 and d

 A

< 0 (volume of A shrinks)

and dS > 0 as it must beIf p

> A^

p^ B

, then

( p

– A^

p^ B

) > 0 and d

 A

> 0 (volume of A grows)

and dS > 0 as it must be

P T S U ^  

What aboutsign of T ???

25

MechanicalEngineering

It Makes Sense! ^ So, our definition^ ^

makes sense

^ is

dimensionally consistent

, and

^ satisfies the requirement that

S is maximized

at

thermal and mechanical equilibrium

^ Next, let’s solve for P independent of T

P T

S

U

   

26

MechanicalEngineering

(^1)     



y x z^

z x y z x y

U S U S

S

U ^ 

^ 

T

U

P T

S

       ^

T T P

U^

S

1 ) 1 (

or

S U

P^

   (^1)  (^) 

S

U

P

Solve for the Pressure ^

Recall form calculus the following rule of partialderivatives: ^ For us:

27

MechanicalEngineering

Interlude ^ We considered two isolated systems and cameup with relationships between

T

,^ S

, &

U

and

p ,

S , &

U

.

^ Simple compressible system ^ Relate changes: ^ Apply fundamental relations: 1

st^ law, 2

nd^

law

and some physical reasoning ^ Obtained:

) , (^

^

US S

d S dU S U

dS

U S

U

P

^ 

S U

T^

28